Derivative of unit step function?

How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it.

I am trying to find the derivative of this:

$v(t) = u(t+1) - 2u(t) + u(t-1)$

$u(t) = 0$ when $t < 0$

$u(t) = 1$ when $t > 0$

The relationship between unit step function and impulse function:

δ(n) = u(n) - u(n-1)

$δ(t)=du(t)/dt$

• Continuous $\displaystyle \delta(t) = \frac{d u(t)}{dt}$ – arthur Nov 1 '16 at 0:15
• Discrete $\displaystyle \delta[n] = u[n] - u[n-1]$ – arthur Nov 1 '16 at 0:17
• Possible duplicate of How to prove that the derivative of Heaviside's unit step function is the Dirac delta?. – user137731 Nov 1 '16 at 0:22
• @arthur Thanks for the clarification. I've edited the post. – zdub Nov 1 '16 at 0:24
• @zdub did you mean for one of the Heaviside functions to be $u(t\color{red}{+}1)$? Otherwise the first and third terms are the same and can be simplified to $2u(t-1)$. – user137731 Nov 1 '16 at 0:26

1 Answer

The derivative of unit step $u(t)$ is Dirac delta function $\delta(t)$, since an alternative definition of the unit step is using integration of $\delta(t)$ here.

$$u(t)=\int_{-\infty}^{t}\delta(\tau)d\tau$$

Hence,

$$\frac{dv}{dt}=\delta(t-1)-2\delta(t)+\delta(t-1)$$

• Thanks! So the graph of the derivative would just be vertical lines at t = -1, 0 and 1? – zdub Nov 1 '16 at 0:21
• @zdub The way that engineers draw it, there should be vertical arrows at $1$ and $0$, each of which is $2$ units tall (but at $0$ it should point in the opposite direction). – user137731 Nov 1 '16 at 0:24
• @zdub You have two $u(t-1)$ in your question, or $2u(t-1)$. This means the derivative is a $2\delta(t-1)$ that is represented by an upward arrow with amplitude $2$ at $t=+1$. If your actual question is $u(t-1)+2u(t)+u(t+1)$, then there will be three upward delta functions in the derivative: two of them are at $t=\pm1$ with unit scale, and the third at origin with scale $2$. – msm Nov 1 '16 at 0:33